大数操作，用顺序表实现最大一百位数的加减法实现

直接贴代码，看注释：

#include 
#include 

// 最大一百位数的加减法实现
const int MAXSIZE = 100;

typedef struct {
	
	int length;						//长度
	int numbers[MAXSIZE];			//各个位上的数字，从低位数字到高位数字
	int sign;						//符号：0代表正数，1代表负数，规定0的符号为正

}SqList;

//头插，即新元素插入到numbers[0]
void insert(SqList *l, int x)
{
	SqList *p = l;
	
	if(l->lengthlength-1;i>=0;i--){	
			l->numbers[i+1]=l->numbers[i];
		}

		l->numbers[0] = x;
		l->length++;
	}
	
}


//初始化： s：符号，0正1负；a：包含从高位到低位各个位数的数组；n：位数，也即数组a的长度
SqList *init(int s, int *a, int n)
{
	SqList *L;
	L=(SqList*)malloc(sizeof(SqList));
	if(!L)
	{
		printf("Init Error!");
		exit(1);
	}
	L->sign=s;
	L->length=0;

	int i;
	for(i=0;ilength);
		printf("sign: %d\n", l->sign);
		printf("numbers: ");
		
		int i=0;
		
		for(i=l->length-1;i>=0;i--){
			printf("%d ", l->numbers[i]);	
		}
		printf("\n");
	}


}

// 简单的正数相加
SqList *add(SqList *l1, SqList *l2){

	SqList *l = init(0,0,0);
	
	int c = 0;		//进位

	int i;
	int t;		//temp value for sequence add.
	
	int len = (l1->length>l2->length)?l2->length:l1->length;
	
	//按照加法逻辑，从低位到高位开始相加
	for(i=0;inumbers[i] + l2->numbers[i] + c;
		
		l->numbers[i] = t%10;
		l->length++;

		if(t>=10){
			c = t/10;
		}else{
			c = 0;
		}
	}

	//按照加法逻辑，以下处理较大数字的高位
	len = l1->length-l2->length;
	len = (len<0)?-len:len;
	len = len+i;

	for(i;ilength>l2->length){ 
			t = l1->numbers[i] + c;
		}else{
			t = l2->numbers[i] + c;
		}

		l->numbers[i] = t%10;
		l->length++;

		if(t>=10){
			c = t/10;
		}else{
			c = 0;
		}
	}

	return l;
}

// 简单的正数相减，不论顺序，永远返回较大数减去较小数的值
SqList *sub(SqList *l1, SqList *l2){

	SqList *l = init(0,0,0);
	int c = 0;		// 借位	

	int i;
	int t;		
	
	int len = (l1->length>l2->length)?l2->length:l1->length;
	
	//按照减法逻辑，从低位到高位开始相减
	for(i=0;ilength>l2->length){
			if((l1->numbers[i]-c)numbers[i]){
				t = l1->numbers[i]+10 - l2->numbers[i] - c;
				c = 1;
			}else{
				t =	l1->numbers[i] - l2->numbers[i] - c;
				c = 0;
			}
				
		}else{

			if((l2->numbers[i]-c)numbers[i]){
				t = l2->numbers[i]+10 - l1->numbers[i] - c;
				c = 1;
			}else{
				t =	l2->numbers[i] - l1->numbers[i] - c;
				c = 0;
			}
			
		}
		
		l->numbers[i] = t%10;
		l->length++;
	
	}

	//处理较大数的高位
	len = l1->length-l2->length;
	len = (len<0)?-len:len;
	len = len+i;

	for(i;ilength>l2->length){ 
			t = l1->numbers[i] - c;
		}else{
			t = l2->numbers[i] - c;
		}

		//如果和借位相减为0，得数应该减少一位
		if(t==0){
			l->length--;
		}

		l->numbers[i] = t%10;
		l->length++;

		if(t<0){
			c = 1;
		}else{
			c = 0;
		}
	}

	return l;
}

//绝对值比较，用于真正的加减运算判断
int cmpabs(SqList* l1, SqList *l2){
	if(l1->length!=l2->length){
		return (l1->length>l2->length)?(1):(0);
	}else{
		int i;
		for(i=l1->length-1;i>=0;i--){
			if(l1->numbers[i]>l2->numbers[i]){
				return 1;
			}else if(l1->numbers[i]numbers[i]){
				return 0;
			}
		}
		return 2;
	}

}

//加法，判断两数符号做相应处理
SqList* ADD(SqList* l1, SqList* l2){
	if(l1->sign == 0 && l2->sign == 0){
		return add(l1,l2);
	}else if(l1->sign == 1 && l2->sign == 1){
		SqList* l = add(l1,l2);
		l->sign = 1;
		return l;
	}else{
		
		SqList* l = sub(l1,l2);
		l->sign = !cmpabs(l1,l2);
		return l;
	}

}

//减法，判断两数符号做相应处理，较加法而言还要考虑操作数顺序
SqList* SUB(SqList* l1, SqList* l2){

	int t = cmpabs(l1,l2);
	if(t == 2){
		int a[1] = {0};
		SqList* l = init(0,a,1);
		return l;
	}

	if(l1->sign == 1 && l2->sign == 1){
		if(t == 0){
			return sub(l1,l2);
		}else if(t == 1){
			SqList* l = sub(l1,l2);
			l->sign = 1;
			return l;
		}

	}else if(l1->sign == 0 && l2->sign == 0){
		
		if(t == 1){
			return sub(l1,l2);
		}else if(t == 0){
			SqList* l = sub(l1,l2);
			l->sign = 1;
			return l;
		}
	}else if(l1->sign == 1 && l2->sign == 0){
		SqList* l = add(l1,l2);
		l->sign = 1;
		return l;
		
	}else if(l1->sign == 0 && l2->sign == 1){
		
		return add(l1,l2);
	}
	
}

int main()
{

    printf("Application starts...\n");
	
	int a1[3] = {1,2,2};
	SqList *l1 = init(1, a1, 3);
	log(l1);

	int a2[2] = {2,4};
	SqList *l2 = init(0, a2 ,2);
	log(l2);

	SqList *l3 = init(1, a2 ,2);

	SqList *l = SUB(l2,l1);
	log(l);

	return 0;

}